For what $n$, is $S_{n}$ a homomorphic image of $S_{n+1}$?

Let's let $\phi\colon S_{n+1}\to S_n$ be a surjective homomorphism. Then, as you said the kernel of $\phi$ must be a normal subgroup of $S_{n+1}$, and $|\text{ker }\phi|=n+1$.

Luckily, the normal subgroups of the symmetric groups are well-understood. Here is a proof that the only normal subgroup of $S_n$ is $A_n$, when $n\geq 5$. (That proof relies on the knowledge that the alternating groups $A_n$ are simple for $n\neq 4$, a proof of which can be found here.) Clearly, for $n+1\geq 5$, the alternating group $A_{n+1}\leq S_{n+1}$ will be far too large to be our kernel; therefore, we only have to check the cases $n=1,2,3$.

$n=1$ is the easiest; $S_1$ is trivial, so the trivial homomorphism $\phi_{id}\colon S_2\to S_1$ works.

For $n=2$, we can easily see that the sign homomorphism, which sends even homomorphisms to the identity and odd homomorphisms to the transpositions $(1\ 2)$, will do nicely as our homomorphism $\phi\colon S_3\to S_2$.

$n=3$ is far more interesting; $A_4$ is actually non-simple, and $S_4$ has a normal subgroup $H$ that is isomorphic to the Klein $4$-group. This subgroup, it turns out, is perfect for our kernel. A description of this homomorphism $\phi:S_4\to S_3$ is left as an exercise to the reader :-)